I can't understand the electrolytic capacitors, when a capacitor has a capacitance of 100 microfarads, does that mean that when it is charged with 100 volts will the charge of the plate be 0.01 coulomb? If there is a part of the plate with no isolation, then I touch it, I will be shocked with a charge of 0.01 coulomb and 100 volts?

1 Answer
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a capacitor has a capacitance of 100 microfarads, does that mean that when it is charged with 100 volts will the charge of the plate be 0.01 coulomb?

Well, it has two plates, with an equal and opposite charge of $0.01 \:\mathrm{C}$ The top cap is connected to one of them.

I will be shocked with a charge of 0.01 coulomb

A fraction of the $0.01 \:\mathrm{C}$ will be discharged into your body. The thing is, the charge is "held" there by the other plate, you can't discharge one plate unless you simultaneously discharge the other. If the other plate is grounded (and you are as well), then yes, you will get a full discharge of $0.01 \:\mathrm{C}$ .

100 volts?

Depends on the configuration. If the other plate of the capacitor is grounded and you touch the cap, yes, you will have a $100\:\mathrm{V}$ potential difference across your body. If some part of the circuit is grounded, you will have a significant p.d. set up across your body. If it's not grounded at all, then there's a chance that the p.d. set up will be negligible.

It all really depends on the rest of the circuit, specifically where it is grounded. To discharge a capacitor you need a closed circuit, so there must be some path that connects the two plates through your body and ground for something significant to happen.